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WBJEE 2019 - Maths
Question 1: P is the extremity of the latus rectum of ellipse 3x2 + 4y2 = 48 in the first quadrant. The eccentric angle of P is
- a. π/8
- b. 3π/4
- c. π/3
- d. 3π/3
Solution:
Answer: c
Given
Equation of ellipse is
\(\frac{x^{2}}{16}+\frac{y^{2}}{12}=1\)Let point P on curve (4cos θ,
\(2\sqrt{3}sin\theta\))But we know point P is extremity of latus rectum of ellipse
So, P is (2, 3)
On comparing point P
4 cos θ= 2 and
\(2\sqrt{3}sin\theta\)=3Cos θ = (1/2) and sin θ = √3/2 ⇒ θ= π/3
Question 2: The direction ratios of the normal of the plane through the points (1,2,3), (–1,–2,1) and parallel to
- a. (2, 3, 4)
- b. (14, –8, –1)
- c. (–2, 0, –3)
- d. (1, –2, –3)
Solution:
Answer: b
Normal of the plane passing through the points (1, 2, 3), (–1, –2, 1) and parallel to given line is
=
\(\begin{vmatrix} \hat{i} &\hat{j} & \hat{k}\\ 2 & 3 & 4\\ 1+1 & 2+2 & -3-1 \end{vmatrix}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 &3 & 4\\ 2 & 4 & -4 \end{vmatrix}\)=
\(-28\hat{i}+16\hat{j}+2\hat{k}\)Direction ratio’s of the normal to plane is < 14, –8, –1>
Question 3: The equation of the plane which bisects the line joining the points (1, 2, 3) and (3, 4, 5) at right angles is
- a. x + y + z = 0
- b. x + y – z = 9
- c. x + y +z = 9
- d. x + y – z + 9 = 0
Solution:
Answer: c
Equation of plane is (3 – 1)x +(4 – 2)y + (5 – 3)z = k …..(1)
This plane passes through m
\(=\left [ \frac{3+1}{2},\frac{2+4}{2},\frac{5+3}{2} \right ]\)= (2, 3, 4)
⇒k = 2 × 2 + 2 × 3 + 2 × 4 = 18 (put in eq (1))
⇒ Eq. of plane is x + y + z =9
Question 4: The limit of the interior angle of a regular polygon of n sides as n → ∞is
- a. π
- b. π/8
- c. 3π/2
- d. 2π/3
Solution:
Answer: a
We know,
One interior angle = (n-2) π/n
\(\lim_{n\rightarrow \infty }\frac{(n-2)\pi }{n}=\pi\)
Question 5: Let f(x) > 0 for all x and f ‘(x)’ exists for all x. If f is the inverse function of h and
- a. 1 + log (f(x))
- b. 1+ f(x)
- c. 1 – log (f(x))
- d. logf(x)
Solution:
Answer: a
Given
⇒f(x) =h-1(x)
⇒h(f(x)) = x
On differentiating with respect to x
⇒h’(f(x)) . f’(x) = 1
⇒ f’(x) = [1/ h’(f(x))] = 1 + log (f(x))
Question 6: Consider the function f(x) = cosx2. Then
- a. f is of period 2π
- b. f is of period √2π
- c. f is not periodic
- d. f is of periodic π
Solution:
Answer: c
Given
f(x) = cosx2
Let, f(x) is periodic with period T
We know, f(x + T) = f(x)
⇒cos (x + T)2 = cos x2
⇒ cos (x + T)2 – cosx2 = 0
⇒
\(-2sin\left ( \frac{(x+T)^{2}-x^{2}}{2} \right )sin\left ( \frac{(x+T)^{2}+x^{2}}{2} \right )=0\)⇒ (x + T)2 – x2 = nπ or (x + T)2 + x2 = nπ
Which is not possible because these equations are quadratic equation not identity
⇒ f(x) is not periodic
Question 7:
- a. Does or exist finitely
- b. is 1
- c. is e2
- d. is 2
Solution:
Answer: c
\(\lim_{x\rightarrow 0^{+}}(e^{x}+x)^{1/x}(1^{\infty })\)\(\lim_{e^{x\rightarrow 0^{+}}}\frac{(e^{x}+x)-1}{x}\)Using binomial expansion of ex =1 + x + x2/2!+…….
⇒ e 1+1 =e2
Question 8: Let f(x) be a derivable function, f ‘(x) > f(x) and f(0) = 0. Then
- a. f(x) > 0 for all x > 0
- b. f(x) < 0 for all x > 0
- c. no sign of f (x) can be ascertained
- d. f(x) is a constant function
Solution:
Answer: a
Given
f ‘(x) > f(x)
Multiply e–x both sides
⇒f’(x). e–x – f(x).e–x> 0
⇒ (f(x). e–x)’> 0
⇒e –x. f(x) is increasing function
⇒
\(e^{-x}.f(x)> e^{-0}.f(0)\forall x> 0\)
⇒e –x. f(x) >0 since, f(0) = 0 (given)
⇒
\(f(x)> 0 \forall x> 0\)
Question 9: Let f: [1, 3] → R be a continuous function that is differentiable in (1, 3) and f ‘(x) = |f(x)|2 + 4 for all x ∈ (1, 3). Then
- a. f(3) – f(1) = 5 is true
- b. f(3) – f(1) = 5 is false
- c. f(3) – f(1) = 7 is false
- d. f(3) – f(1) < 0 only at one point (1, 3)
Solution:
Answer: (b,c)
f ‘(x) = |f(x)|2 + 4
Using LMVT theorem
⇒
\(\frac{f(3)-f(1)}{3-1}\)= f’ (c) for atleast one c∈ (1, 3)⇒f(3) – f(1) = 2(f(c))2 + 8
⇒f(3) – f(1) ≥ 8
⇒f(3) – f(1) = 5 (false)
Similarly f(3) – f(1) = 7 (false)
Question 10:
- a. does not exist
- b. exists and is zero
- c. exists and is 1
- d. exists and is e–1
Solution:
Answer: b
\(\lim_{x\rightarrow 0^{+}}(x^{n}lnx),n> 0\)\(\lim_{x\rightarrow 0^{+}}\left [ \frac{lnx}{1/x^{n}} \right ]\left [ \frac{\infty }{\infty } \right ]\)formUsing L-Hospital Rule
\(\lim_{x\rightarrow 0^{+}}\left [ \frac{1/x}{\frac{-n}{x^{n+1}}} \right ]\)\(\lim_{x\rightarrow 0^{+}}\left [ \frac{x^{n}}{-n} \right ]=0\)
Question 11: If
- a. c
- b. c – x
- c. c + x
- d. 2x + c
Solution:
Answer: b
Question 12:
- a. straight lines
- b. circles
- c. ellipses
- d. parabolas
Solution:
Answer: d
Question 13: The value of the integration
- a. is independent of λ only
- b. is independent of μ only
- c. in independent of γ only
- d. depends on γ, μ and λ
Solution:
Answer: b
Question 14: The value of
- a. \(e^{sin^{2}y}\)
- b. e2siny
- c. \(e^{\left | siny \right |}\)
- d. \(e^{cosec^{2}y}\)
Solution:
Answer: a
Question 15: If
- a. 1/log x
- b. log 2. c. (log2)2
- d. 1/(log2)2
Solution:
Answer: d
\(\int 2^{2^x}.2^{x}dx=A2^{2^x}+c\)---------(1)Put 22x = t ⇒(22^x.ln2)(2x.ln2)dx = dt
\(\Rightarrow \int \frac{dt}{(ln2)^{2}}\)\(\Rightarrow \frac{t}{(ln2)^{2}}+C\Rightarrow \frac{2^{2x}}{ln2^{2}}+C\)(on comparing with equation (1))
⇒A = 1/(ln2)2
Question 16: The value of the integral
- a. 0
- b. 1 – e–1
- c. 2e–1
- d. 2(1 – e–1)
Solution:
Answer: d
Question 17:
- a. does not exist
- b. is 1
- c. is 2
- d. is 3
Solution:
Answer: c
= (2/3) x 3 =2
Question 18: The general solution of the differential equation
- a. x-ye(x/y) = c
- b. y-xe(x/y) =c
- c. x+ye(x/y)= c
- d. y+xe(x/y) = c
Solution:
Answer: c
Question 19: General solution of
- a. (x/a) = tan(y/a) +c
- b. tan xy = c
- c. tan (x+y)= c
- d. tan(y+c/a) = (x+y/a)
Solution:
Answer: d
\((x+y)^{2}\frac{dy}{dx}=a^{2}\Rightarrow \frac{dy}{dx}=\frac{a^{2}}{(x+y)^{2}}\)Put x + y = t ⇒ 1 + (dy/dx) = (dt/dx) …….(2)
⇒ From (1) & (2)
⇒
\(\Rightarrow dx=\frac{t^{2}}{t^{2}+a^{2}}dt\)\(\Rightarrow dx=\left [ 1-\frac{a^{2}}{t^{2}+a^{2}} \right ]dt\)Integrating both sides
⇒x + c = t – a tan–1 t/a since, t = x + y
⇒x = x + y – a tan–1(x+y/a) +c
\(\Rightarrow atan^{-1}\left ( \frac{x+y}{a} \right )=y+c\Rightarrow \frac{x+y}{a}=tan\left (\frac{y+c}{a} \right )\)
Question 20. Let P(4, 3) be a point on the hyperbola (x2/a2)-(y2/b2)=1.If the normal at P intersects the x-axis at (16, 0), then the eccentricity of the hyperbola is
- a. \(\frac{\sqrt{5}}{2}\)
- b. 2
- c. \(\sqrt{2}\)
- d. \(\sqrt{3}\)
Solution:
Answer: b
Question 21: If the radius of a spherical balloon increases by 0.1% then its volume increases approximately by
- a. 0.2%
- b. 0.3%
- c. 0.4%
- d. 0.05%
Solution:
Answer: b
Volume of spherical balloon V = (4/3)πr3
\(\begin{aligned} \frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=\frac{\left(\frac{4}{3} \pi\left(\mathrm{r}+\frac{\mathrm{r}}{10}\right)^{3}-\frac{4}{3} \pi \mathrm{r}^{3}\right)}{\frac{4}{3} \pi^{3}} \times 100 \\ &=\left(1+\frac{1}{10}\right)^{3}-1 \end{aligned}\)= 1 +(3/10) +(3/100) +(1/1000) –1 ≈ 0.3% approximately
Question 22: The three sides of a right-angled triangle are in G.P. (Geometrical Progression). If the two acute angles be α and β, then tan α and tan β are:
- a. \(\frac{\sqrt{5}+1}{2} and \frac{\sqrt{5}-1}{2}\)
- b. \(\sqrt{\frac{\sqrt{5}+1}{2}} and \sqrt{\frac{\sqrt{5}-1}{2}}\)
- c. \(\sqrt{5}and \frac{1}{\sqrt{5}}\)
- d. \(\frac{\sqrt{5}}{2}and \frac{2}{\sqrt{5}}\)
Solution:
Answer: b
Let sides are a, ar, ar2 (r > 1)
Using Pythagoras theorem
⇒a2 + a2r2 = a2 r4
⇒r4 + r2–1 = 0
\(\Rightarrow r^{2}=\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}\)(not possible)\(r=\sqrt{\frac{-1+\sqrt{5}}{2}}\)\(tan\alpha =\frac{1}{r}\)\(=\sqrt{\frac{2}{\sqrt{5}-1}}=\sqrt{\frac{\sqrt{5}+1}{2}}\)\(tan\beta =r=\sqrt{\frac{\sqrt{5}-1}{2}}\)
Question 23: If
- a. (1/4,1/3)
- b. (1/4 ,1/2)
- c. (-1/4 ,1/2)
- d. (1/3 ,-1/2)
Solution:
Answer: b
\(log_{2}6+\frac{1}{2x}=log_{2}(2^{1/2}+8)\)\(log_{2}\left [ \frac{2^{1/x}+8}{6} \right ]= \frac{1}{2x}\)On taking anti log
\(\Rightarrow \frac{2^{1/x}+8}{6}=2^{1/2x}\)\(\Rightarrow \left [ 2^{\frac{1}{2x}} \right ]^{2}-6.2^{1/2x}+8=0\)⇒t2-6t+8 =0
⇒ t = 4,2
⇒ 2(1/2x) = 4,2
⇒ 2x = 1, ½
⇒ x = ½,1/4
Question 24: Let z be a complex number such that the principal value of argument, arg z > 0. Then arg z – arg(–z) is
- a. π/2
- b. ± π
- c. π
- d.– π
Solution:
Answer: c
⇒argz – arg(–z)
⇒argz – [argz – arg(–1)] since, arg (z1/z2) =argz1 – argz2
⇒ argz – [argz – arg(–1)] since, arg(–1) = π
⇒ π
Question 25: The general value of the real angle θ, which satisfies the equation
(cos θ + i sin θ)(cos2 θ + i sin2 θ)..........(cos n θ + i sin θ) = 1 is given by (assuming k is an integer)
- a. \(\frac{2k\pi }{n+2}\)
- b. \(\frac{4k\pi }{n(n+1)}\)
- c. \(\frac{4k\pi }{(n+1)}\)
- d. \(\frac{6k\pi }{n(n+1)}\)
Solution:
Answer: b
⇒ (cos θ + i sin θ) (cos 2θ + i sin 2θ)…………(cos nθ + i sin nθ) = 1
⇒ e i θ.e i2 θ………e in θ =1 (using Euler’s formula)
⇒ e i θ(1+2+………+n) =1
\(e^{i\theta \frac{n(n+1)}{2}}=1\)⇒
\(\left [ \frac{n(n+1)\theta }{2} \right ]=2k\pi\)⇒
\(\theta =\frac{4k\pi }{n(n+1)}\)
Question 26: Let a,b,c be real numbers such a + b + c < 0 and the quadratic equation ax2 + bx + c = 0 has imaginary roots. Then
- a. a > 0, c > 0
- b. a > 0, c < 0
- c. a < 0, c > 0
- d. a < 0, c < 0
Solution:
Answer: d
Given
⇒ ax2 + bx + c = 0 has imaginary roots ⇒D < 0
⇒ a + b + c < 0 ⇒f(1) < 0
⇒ f(x) < 0
\(\forall\)x∈R and a < 0⇒ f(0) < 0 ⇒ c < 0
Question 27: A candidate is required to answer 6 out of 12 questions which are divided into two parts A and B each containing 6 questions and he/she is not permitted to attempt more than 4 questions from any part. In how many different ways can he/she make up his/her choice of 6 questions?
- a. 850
- b. 800
- c. 750
- d. 700
Solution:
Answer: a
⇒6C2 × 6C4 + 6C3× 6C3 + 6C4 × 6C2
⇒ (15)2 + (20)2 + (15)2
⇒225 + 400 + 225 = 850
Question 28: There are 7 greetings cards, each of a different colour and 7 envelopes of same 7 colours as that of the cards. The number of ways in which the cards can be put in envelopes, so that exactly 4 of the cards go into envelopes of respective colour is:
- a. 7C3
- b. 2. 7C3
- c. 3! 4C4
- d. 3! 7C34C3
Solution:
Answer: b
⇒7C4 × (De-arrangement of 3 things)
⇒ 35 × 2 = 70 = 7C3× 2
Question 29.72n + 16n – 1 (n ∈ N) is divisible by
- a. 65
- b. 63
- c. 61
- d. 64
Solution:
Answer: d
⇒72n + 16n –1
⇒ (8 – 1)2n + 16n– 1
⇒ [2nC082n _ 2nC182n-1 + --------2nC 2n-181+2nC2n]+16n-1
⇒[64
\(\lambda\)-2n.8+1] + 16n – 1⇒64
\(\lambda\)– 16n + 1 + 1 6n – 1⇒64
\(\lambda\)
Question 30: The number of irrational terms in the expansion of (3 1/8 + 5 1/4)84 is
- a. 73
- b. 74
- c. 75
- d. 76
Solution:
Answer: b
⇒ (84-n)/8 = rational⇒ n = 4, 12 ……………, 76
⇒n/4= rational ⇒ n = 4, 8, 12...……………., 84
⇒ n can take total 11 terms
⇒Total number of rational terms = 11
Irrational terms = total terms – rational terms = 85 – 11 = 74
Question 31: Let A be a square matrix of order 3 whose all entries are 1 and let I3 be the identity matrix of order 3. Then the matrix A – 3I3 is
- a. invertible
- b. orthogonal
- c. non-invertible
- d. real skew symmetric matrix
Solution:
Answer: c
\(A=\begin{vmatrix} 1 & 1 & 1\\ 1& 1 &1 \\ 1 & 1 &1 \end{vmatrix};I_{3}=\begin{vmatrix} 1 & 0 & 0\\ 0&1 &0 \\ 0& 0 &1 \end{vmatrix}\)\(\Rightarrow A-3I_{3}=\begin{vmatrix} 1 & 1 & 1\\ 1& 1 &1 \\ 1 & 1 &1 \end{vmatrix}-\begin{vmatrix} 3 & 0 & 0\\ 0&3 &0 \\ 0& 0 &3 \end{vmatrix}=\begin{vmatrix} -2 &1 &1 \\ 1& -2 &1 \\ 1& 1 &-2 \end{vmatrix}\)⇒det. (A –3I3) = –2(3) –1(–3) + 1(3) = 0
⇒ A – 3I3 is non invertible matrix
Question 32: If M is any square matrix of order 3 over R and If M’ be the transpose of M, then adj(M’) – adj(M)’ is equal to
- a. M
- b. M’
- c. null matrix
- d. identity matrix
Solution:
Answer: c
We know, for square matrix adj(M’) = (adj M)’
So, adj (M’) – (adj M)’ = 0 = null matrix
Question 33: If
- a. 1/5
- b. 5
- c. 52
- d. 1
Solution:
Answer: a
\(\Rightarrow A=\begin{vmatrix} 5 & 5x & x\\ 0& x &5x \\ 0 & 0 &5 \end{vmatrix}(given)\)⇒
\(\left | A^{2} \right |=25(given)\)⇒
\(\left | A \right |=\pm 5(given)\)\(\Rightarrow \left | A \right |=\begin{vmatrix} 5 & 5x & x\\ 0& x &5x \\ 0 & 0 &5 \end{vmatrix}=\pm 5\)⇒
\(25x=\pm 5\)⇒
\(x=\pm 1/5\)⇒
\(\left |x \right |= 1/5\)
Question 34: Let A and B be two square matrices of order 3 and AB = O3, where O3 denotes the null matrix of order 3. Then
- a. must be A = O3, B = O3
- b. if A \(\neq\)O3, must be B\(\neq\)O3
- c. if A = O3, must be B \(\neq\)O3
- d. may be A \(\neq\)O3, B\(\neq\)O3
Solution:
Answer: d
Let
\(A=\begin{vmatrix} 1 & 0 & 0\\ 0 &0 &0 \\ 0& 0 &0 \end{vmatrix} and B=\begin{vmatrix} 0 &0 &0 \\ 0 & 0 &0 \\ 0 & 0 &1 \end{vmatrix}\)Hence AB = 0 but A
\(\neq\)0, B\(\neq\)0So, if AB = 0 then may be A
\(\neq\)0, B = 0
Question 35: Let P and T be the subsets of X-Y plane defined by
P = {(x,y): x > 0, y > 0 and x2 + y2 = 1}
T = {(x,y): x > 0, y > 0 and x8 + y8< 1}
Then
- a. the void set \(\phi\)
- b. P
- c. T
- d. P–TC
Solution:
Answer: b
Let (h, k) satisfies x2 + y2 = 1 then h2 + k2 = 1
Now h8 + k8 = h8 + (1 – h2)4 = 2h8 –4h6 – 4h2 + 1
= 2h2 (h2 – 1) (h4 – h2 + 2) + 1
= –2h2k2 (h4 –h2 + 2) + 1 < 1
\(\forall\)h > 0, k > 0All solution of x2 + y2 = 1 satisfies x8 + y8< 1
⇒
\(P\cap T\)=P
Question 36: Let f: R
- a. f is one-one but not onto mapping
- b. f is onto but not one-one mapping
- c. f is both one-one and onto
- d. f is neither one-one nor onto
Solution:
Answer: d
\(f(x)=x^{2}-\frac{x^{2}}{1+x^{2}}\)\(\Rightarrow f(x)=\frac{x^{4}}{1+x^{2}}\)Therefore, f(-x) =f(x)
Therefore, f(x) is even function, hence it is many one.
Also f(x) ≥ O
\(\forall\)x\(\epsilon\)R , hence it is into functionTherefore, f(x) is neither one-one nor onto
Question 37: Let the relation ρ be defined on R as a ρb if 1 + ab > 0. Then
- a. ρ is reflexive only
- b. ρ is an equivalence relation
- c. ρ is reflexive and transitive but not symmetric
- d. ρ is reflexive and symmetric but not transitive
Solution:
Answer: d
(a, a)
\(\epsilon\)because 1 + a2> 0 ⇒ ρ is reflexiveIf 1 + ab > 0 then 1 + ba > 0 ⇒ if (a, b)
\(\epsilon\)ρ then (b, a)\(\epsilon\)ρ⇒ ρ is symmetric
Now (-4, 1/8)
\(\epsilon\)ρ and (1/8,5)\(\epsilon\)ρBut (–4, 5) ∉ ρ, hence ρ is not transitive
Question 38: A problem in mathematics is given to 4 students whose chances of solving individually are 1/2,1/3 ,1/4 and 1/5. Then probability that the problem will be solved at least by one student is
- a. 2/3
- b. 3/5
- c. 4/5
- d. ¾
Solution:
Answer: c
Probability that no student solves problem is
=
\(\left [ 1-\frac{1}{2} \right ]\left [ 1-\frac{1}{3} \right ]\left [ 1-\frac{1}{4} \right ]\left [ 1-\frac{1}{5} \right ]=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}=\frac{1}{5}\)⇒ Probability that the problem will be solved by at least one student is = 1 – (1/5) = 4/5
Question 39: If X is a random variable such that σ (X) = 2.6, then σ (1 – 4X) is equal to
- a. 7.8
- b. –10.4
- c. 13
- d. 10.4
Solution:
Answer: d
Given
σ (x) = 2.6
We know
σ (ax + b) = |a| (σ (x))
So s(–4x + 1) = |–4|( σ (x)) = 4 × 2.6 = 10.4
Question 40: If esinx – e–sinx – 4 = 0, then the number of real values of x is
- a. 0
- b. 1
- c. 2
- d. 3
Solution:
Answer: a
esinx – e–sinx = 4
Let esinx = t
⇒ t2 –4 t –1 = 0
⇒
\(t=2+\sqrt{5},2-\sqrt{5}\)We know that t is real positive number
⇒
\(t=2+\sqrt{5}\)⇒
\(e^{sinx}=2+\sqrt{5}\)⇒ [e-1,e1] ∉
\(2+\sqrt{5}\)⇒ hence no solution exist
Question 41: The angles of a triangle are in the ratio 2 : 3 : 7 and the radius of the circumscribed circle is 10 cm. The length of the smallest side is
- a. 2 cm
- b. 5 cm
- c. 7cm
- d. 10 cm
Solution:
Answer: d
Let, angles are 2x, 3x, 7x
We know
⇒2x + 3x + 7x = 180° ⇒ 12x = 180° ⇒ x = 15°
⇒ angles are – 30°, 45°, 75°
⇒ length of smallest side a
⇒a/sin A=2R
⇒a = 2Rsin A
⇒a = 2 × 10 × sin 30° = 10
Question 42: A variable line passes through a fixed point (x1, y1) and meets the axes at A and B. If the rectangle OAPB be completed, the locus of P is, (O being the origin of the system of axes)
- a. (y – y1)2 = 4(x – x1)
- b. (x1/x) + (y1/y) = 1
- c. x2 + y2 = x12 + y12
- d. \(\frac{x^{2}}{2x_{1}^{2}}+\frac{y^{2}}{2y_{1}^{2}}=1\)
Solution:
Answer: b
Let P(h, k) then A is (h, 0) & B is (0, k)
Equation of AB is
\(\frac{x}{h}+\frac{y}{k}=1\)which passes through (x1, y1) is⇒
\(\frac{x_{1}}{h}+\frac{y_{1}}{k}=1\)⇒ Required locus is
\(\frac{x_{1}}{x}+\frac{y_{1}}{y}=1\)
Question 43: A straight line through the point (3, -2) is inclined at an angle 60° to the line x + y = 1. If it intersects the X-axis, then its equation will be
- a. \(y+x\sqrt{3}+2+3\sqrt{3}=0\)
- b. \(y-x\sqrt{3}+2+3\sqrt{3}=0\)
- c. \(y-x\sqrt{3}-2-3\sqrt{3}=0\)
- d. \(x-x\sqrt{3}+2-3\sqrt{3}=0\)
Solution:
Answer: b
L1:
\(\sqrt{3}x+y=1\Rightarrow m_{1}=\sqrt{3}\)θ = 60° (given)
\(\Rightarrow tan\theta =\left | \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |\)⇒
\(\pm \sqrt{3}=\frac{-\sqrt{3}-m_{2}}{1-\sqrt{3}m_{2}}\)⇒ m2 =√3,0 (not possible because lines are not parallel)
⇒m2 =√3
So, equation of line passing through (3, –2) and the slope √3 is
⇒ y + 2 =√3 (x – 3)
⇒y-√3x+2+3√3=0
Question 44: A variable line passes through the fixed point (α, β). The locus of the foot of the perpendicular from the origin on the line is
- a. x2 + y2 – αx – βy = 0
- b. x2 – y2 + 2αx + 2βy = 0
- c. \(ax+bx\pm \sqrt{(\alpha ^{2}+\beta ^{2})}=0\)
- d. \(\frac{x^{2}}{\alpha ^{2}}+\frac{y^{2}}{\beta ^{2}}=1\)
Solution:
Answer: a
MPQ • MOP = –1
⇒
\(\left ( \frac{k-\beta }{h-\alpha } \right )\left ( \frac{k}{h} \right )=-1\)⇒k2 – kβ = – (h2 –hα)
⇒ h2 + k2 –hα– kβ = 0
⇒Required locus is x2 + y2 –α x –β y = 0
Question 45: If the point of intersection of the lines 2ax + 4ay + c = 0 and 7bx + 3by – d = 0 lies in the 4th quadrant and is equidistant from the two axes, where a, b, c and d are non-zero numbers, then ad: bc equals to
- a. 2 : 3
- b. 2 : 1
- c. 1 : 1
- d. 3 : 2
Solution:
Answer: b
Let point on 4th quadrant which is equidistant from both the axis is (α, –α)
⇒ L1: 2aα – 4a α + c = 0 ⇒α=c/2a …………(1)
⇒L2: 7bα – 3bα – d = 0⇒ α = d/4b …………..(2)
From equation (1) & (2)
⇒ (c/2a) = (d/4b)
⇒ (4/2) = ad/cb
⇒ ad : bc = 2 : 1
Question 46: A variable circle passes through the fixed point A(p, q) and touches x-axis. The locus of the other end of the diameter through A is
- a. (x – p)2 = 4qy
- b. (x – q)2 = 4py
- c. (y – p)2 = 4qx
- d. (y – q)2 = 4px
Solution:
Answer: a
Let other end is (h, k) then centre equal to
\(\left ( \frac{p+h}{2},\frac{q+k}{2} \right )\)Because circle touches x-axis hence radius =
\(\left | \frac{q+k}{2} \right |\)We know
⇒ (AB) Diameter = 2r
⇒
\(\sqrt{(h-P)^{2}+(k-q)^{2}}=2\left | \frac{q+k}{2} \right |\)
On squaring both sides
⇒ (h – p)2 + (k – q)2 = (q + k)2
⇒ (h – p)2 = (k + q)2 – (k – q)2
⇒ Required locus is = (x – p)2 = 4qy
Question 47: If P(0, 0), Q(1, 0) and R
- a. \(\left [ \frac{1}{2},\frac{1}{4} \right ]\)
- b. \(\left [ \frac{1}{2},\frac{\sqrt{3}}{4} \right ]\)
- c. \(\left [ \frac{1}{2},\frac{1}{2\sqrt{3}} \right ]\)
- d. \(\left [ \frac{1}{2},\frac{-1}{\sqrt{3}} \right ]\)
Solution:
Answer: c
ΔPQR is equilateral triangle
So, incentre is same as centroid
⇒ incentre = centroid = centre of circle =
\(\left [ \frac{1+\frac{1}{2}+0}{3},\frac{0+0+\frac{\sqrt{3}}{2}}{3} \right ]=\left [ \frac{1}{2},\frac{1}{2\sqrt{3}}, \right ]\)
Question 48: For the hyperbola
- a. Directrix
- b. Vertices
- c. foci
- d. Eccentricity
Solution:
Answer: c
Focus are =
\((\pm \sqrt{a^{2}+b^{2},0})=(\pm \sqrt{cos^{2}\alpha +sin^{2}\alpha },0)\)⇒ focus are (±1,0), which is independent of α
⇒focus are fixed
Question 49: S and T are the foci of an ellipse and B is the end point of the minor axis. If STB is equilateral triangle, the eccentricity of the ellipse is
- a. 1/4
- b. 1/3
- c. 1/2
- d. 2/3
Solution:
Answer: c
S = (ae, 0)
T = (–ae, 0)
B =(0,b)
⇒ (ST)2 = (SB)2
⇒ (2ae)2 = (ae)2 + b2
⇒3(ae)2 = b2
⇒b2 = 3(a2 – b2)
⇒
\(\frac{b^{2}}{a^{2}}=\frac{3}{4}\)⇒
\(e=\sqrt{1-\frac{b^{2}}{a^{2}}}\)⇒
\(e=\sqrt{1-\frac{3}{4}}\)e=1/2
Question 50: The equation of the directrices of the hyperbola 3x2 – 3y2 – 18x + 12y + 2 = 0 is
- a. \(x=3\pm \sqrt{\frac{13}{6}}\)
- b. \(x=3\pm \sqrt{\frac{6}{13}}\)
- c. \(x=6\pm \sqrt{\frac{13}{3}}\)
- d. \(x=6\pm \sqrt{\frac{3}{13}}\)
Solution:
Answer: a
Equation of hyperbola is 3(x2 – 6x) – 3 (y2 –4y) + 2 = 0
⇒ (x – 3)2 – (y – 2)2 = 13/3
⇒
\(\Rightarrow \frac{(x-3)^{2}}{\sqrt{\frac{13}{3}}}-\frac{(y-2)^{2}}{\sqrt{\frac{13}{3}}}=1\Rightarrow e=\sqrt{2}\)We know equation of directrix is X =
\(\pm \frac{a}{e}\)⇒
\(x-3=\pm \frac{\sqrt{13/3}}{\sqrt{2}}\)⇒
\(x=3\pm \frac{\sqrt{13/3}}{\sqrt{2}}\)
Question 51: The graphs of the polynomial x2 – 1 and cos x intersect
- a. at exactly two points
- b. at exactly 3 points
- c. at least 4 but at finitely many points
- d. at infinitely many points
Solution:
Answer: a
y = x2 – 1 & y = cos x
intersect at exactly two points
Question 52: A point is in motion along a hyperbola y = 10/x so that its abscissa x increases uniformly at a rate of 1 unit per second. Then, the rate of change of its ordinate, when the point passes through (5, 2)
- a. increases at the rate of 1/2 unit per second
- b. decreases at the rate of 1/2 unit per second
- c. decreases at the rate of 2/5 unit per second
- d. increases at the rate of 2/5 unit per second
Solution:
Answer: c
Given
(dx/dt) = 1 unit per second
y = 10/x ………(1)
differentiating with respect to x
⇒Ordinate decreases at rate unit per second.
Question 53: Let a = min{x2 + 2x + 3: x
- a. \(\frac{2^{n+1}-1}{3.2^{n}}\)
- b. \(\frac{2^{n+1}+1}{3.2^{n}}\)
- c. \(\frac{4^{n+1}-1}{3.2^{n}}\)
- d. (1/2)(2n – 1)
Solution:
Answer: c
a = min [x2 + 2x + 3; x
\(\epsilon\)R]a = min [(x+1)2+2]
⇒ a = 2
\(b=\lim_{\theta \rightarrow 0}\frac{1-cos\theta }{\theta ^{2}}=\frac{1}{2^{2}}.\frac{2sin^{2}\frac{\theta }{2}}{\frac{\theta ^{2}}{2^{2}}}=\frac{1}{2}\)⇒b=1/2
Now,
⇒
\(\sum_{r=0}^{n}a^{r}b^{n-r}=a^{0}b^{n}+ab^{n-1}+a^{2}b^{n-2}+.........+a^{n}b^{0}\)\(=\left ( \frac{1}{2} \right )^{n}+\frac{2}{2^{n-1}}+\frac{2^{2}}{2^{n-2}}+.........+2^{n}\)\(=\left ( \frac{1}{2} \right )^{n}\left [ 1+4+4^{2}+.......+4^{n} \right ]\)\(\frac{1}{2^{n}}\left [ \frac{4^{n+1}-1}{4-1} \right ]\)\(\left [ \frac{4^{n+1}-1}{3.2^{n}} \right ]\)
Question 54: Let a > b >0 and I(n) = a1/n – b1/n, J(n) = ((a – b)1/n for all n ≥ 2. then
- a. I(n) < J(n)
- b. I(n) > J(n)
- c. I(n) = J(n)
- d. I(n) + J(n) = 0
Solution:
Answer: a
Given
⇒a > b > 0
⇒ a > b
⇒ (b/a) <1
= I(n) < J(n)
Question 55: Let
- a. 5π/6
- b. π/6
- c. π/3
- d. 2π/3
Solution:
Answer: d
\(\left |\hat{\alpha } \right |= \left | \hat{\beta } \right |=\left | \hat{\gamma }\right |=1 (given)\)Now,
\(\hat{\alpha }\times (\hat{\beta }\times \hat{\gamma })=\frac{1}{2}(\hat{\beta }\times \hat{\gamma })\)On comparing both sides
⇒
\(\Rightarrow -\hat{\alpha }.\hat{\beta }=\frac{1}{2}\)⇒
\(\Rightarrow \hat{\alpha }.\hat{\beta }=\frac{-1}{2}\)⇒
\(\left | \hat{\alpha } \right | .\left | \hat{\beta } \right |cos\theta =\frac{-1}{2}\)⇒
\(\theta =\frac{2\pi }{3}\)
Question 56: The position vectors of the points A, B, C and D are
- a. 0
- b. 1
- c. 2
- d. – 4
Solution:
Answer: d
Given
A, B, C, & D are on a plane
Since,
\(\vec{AB}, \vec{AC},\vec{AD}\)are coplanarSince,
\([\vec{AB},\vec{AC},\vec{AD}]=0\)⇒
\(\Rightarrow \begin{vmatrix} 1 & 1 & -3\\ -2& -1 &-3 \\ -1& -1 & -1-\lambda \end{vmatrix}=0\)\(R_{3}\rightarrow R_{3}+R_{1}\)\(\Rightarrow \begin{vmatrix} 1 & 1 & -3\\ -2& -1 &-3 \\ 0& 0 & -4-\lambda \end{vmatrix}=0\)⇒ (-4-λ) (-1 + 2) = 0
⇒λ = -4
Question 57: A particle starts at the origin and moves 1 unit horizontally to the right and reaches P1, then it moves ½ unit vertically up and reaches P2, then it moves 1/4 unit horizontally to right and reaches P3, then it moves 1/8 unit vertically down and reaches P4, then it moves 1/16 unit horizontally to right and reaches P5 and so on. Let Pn = (xn, yn) and xn =α and yn = β. Then (α,β) is
- a. (2, 3)
- b. \(\left [ \frac{4}{3},\frac{2}{5} \right ]\)
- c. \(\left [ \frac{2}{5},1 \right ]\)
- d. \(\left [ \frac{4}{3},3 \right ]\)
Solution:
Answer: b
Question 58: For any non-zero complex number z, the minimum value of |z| + |z – 1| is
- a. 1
- b. 1/2
- c. 0
- d. 3/2
Solution:
Answer: a
Using inequality |A| + |B|≥ |A – B |
⇒|z| +|z – 1|≥ |z –(z – 1)|
⇒ |z| + |z–1| ≥ 1
⇒ minimum value of |z| + |z – 1| is 1
Question 59: The system of equations
λx + y + 3z = 0
2x + µy – z = 0
5x + 7y + z = 0
Has infinitely many solutions in R. Then,
- a. λ = 2, µ = 3
- b. λ = 1, µ = 2
- c. λ = 1, µ = 3
- d. λ = 3, µ = 1
Solution:
Answer: c
For infinitely many solution Δ= 0
\(\Rightarrow\begin{vmatrix} \lambda &1 &3 \\ 2 & \mu & -1\\ 5& 7 &1 \end{vmatrix}=0\)λ μ + 7 λ –7 + 42 – 15 μ =0
⇒ (λ – 15) (μ + 7)+ 140 =0
Now check from options
⇒ (λ, μ) = (1, 3)
Question 60: Let f: X
- a. f–1(A) – f–1(B) \(\supset\)f–1(A – B) but the opposite does not hold
- b. f–1(A) – f–1(B) \(\subset\)f–1(A – B) but the opposite does not hold
- c. f–1(A – B) = f–1(A) – f–1(B)
- d. f–1(A – B) = f–1(A) \(\cup\)f–1(B)
Solution:
Answer: c
We can see that pre-image of A – B i.e. f–1(A – B) will be f–1(A) – f–1(B)
Therefore, f–1 (A – B) = f–1(A) – f–1(B)
Question 61: Let S, T, U be three non-void sets and f : S
- a. g and f are both surjective
- b. g is surjective, f may not be so
- c. f is surjective, g may not be so
- d. f and g both may not be surjective
Solution:
Answer: b
Obvious g is surjective otherwise gof cannot be surjective but there is no need of “f” to be surjective.
Example.
Hence f(x) is not surjective still gof is surjective
Question 62: The polar coordinate of a point P is
- a. \(\left [ 2,\frac{\pi }{4} \right ]\)
- b. \(\left [ 2,\frac{\pi }{6} \right ]\)
- c. \(\left [ -2,\frac{\pi }{4} \right ]\)
- d. \(\left [ -2,\frac{\pi }{6} \right ]\)
Solution:
Answer: a
line joining PQ is bisected by x –axis
So point Q is
\(\left [ 2,\frac{\pi }{4} \right ]\)
Question 63: The length of conjugate axis of a hyperbola is greater than the length of transverse axis. Then the eccentricity e is
- a. = √2
- b. > √2
- c. < √2
- d.1/√2
Solution:
Answer: b
b > a (given)
On squaring both sides
\(\Rightarrow \frac{b^{2}}{a^{2}}> 1\)On adding 1 in both sides
⇒
\(1+\frac{b^{2}}{a^{2}}> 2\)Taking root both sides
\(\Rightarrow \sqrt{1+\frac{b^{2}}{a^{2}}}> \sqrt{2}\)\(\Rightarrow e> \sqrt{2}\)
Question 64: The value of is
- a. \(\frac{\left [ q \right ]}{p}\)
- b. 0
- c. 1
- d. ∞
Solution:
Answer: a
Question 65: Let f(x) = x4 – 4x3 + 4x2 + c, c
- a. f(x) has infinitely many zeroes in (1, 2) for all c
- b. f(x) has exactly one zero in (1, 2) if –1 < c < 0
- c. f(x) has double zeroes in (1, 2) if –1 < c < 0
- d. whatever be the value of c, f(x) has no zero in (1, 2)
Solution:
Answer: b
f(x) = x4 – 4x3 + 4x2 + c,
Using IVT theorem
For atleast one root/zero = f(1) . f(2) < 0
= (1 + c) c< 0
= c
\(\epsilon\)(–1, 0)
f(x) = x2(x – 2)2 + c
⇒f(x) has exactly one zero in (1, 2) if (–1, 0)
Question 66: Let f and g be differentiable on the interval I and let a, b
- a. If f(a) = 0 = f(b), the equation f'(x) + f(x)g'(x) = 0 is solvable in (a, b)
- b. If f(a) = 0 = f(b), the equation f'(x) + f(x)g'(x) = 0 may not be solvable in (a, b)
- c. If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 is solvable in (a, b), k \(\epsilon\)R
- d. If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 may not be solvable in (a, b), k \(\epsilon\)R
Solution:
Answer: (a,c)
f(a) = 0 = f(b) ⇒f’(a)f’ (b) < 0
Let h(x) = f’ (x) + f(x) g’ (x) …………..(1)
Put x = a in equation (1)
h(a) = f’(a) since f(a) = 0
Put x = b in equation (1)
h(b) = f’ (b) since f(b) = 0
Therefore, h(a) . h(b) < 0 ⇒ h(x) = 0 has roots between (a, b)
Similarly, g’ (x) + kg(x) = 0 has roots between (a, b) as g(a) = 0 = g(b)
Question 67: Consider the function f(x) = (x3/4) – sin πx + 3
- a. f(x) does not attain value within the interval [–2, 2]
- b. f(x) takes on the value 2 in the interval [–2, 2]
- c. f(x) takes on the value 3 in the interval [–2, 2]
- d. f(x) takes no value p, 1 < p < 5 in the interval [–2, 2]
Solution:
Answer:(b,c)
\(f(x)=\frac{x^{3}}{4}-sin\pi x+3\)f(–2) = 1
f(2) = 5
since function is continuous
By intermediate value theorem, f(x) takes all values between 1 to 5
Question 68:Let
- a. a1, a2, a3 are in G.P.
- b. b1, b2, b3 are in A.P.
- c. c1, c2, c3 are in H.P.
- d. a1, a2, a3 are in A.P
Solution:
Answer: (b,d)
\(I_{n}=\int_{0}^{1}x^{n}tan^{-1}xdx\)Using Integration by parts
\(\Rightarrow {I}_{n}=\left.\tan ^{-1} {x} \frac{{x}^{{n}+1}}{n+1}\right|_{0} ^{1}-\int_{0}^{1} \frac{{x}^{n+1}}{n+1} \cdot\left(\frac{1}{1+{x}^{2}}\right) dx\)\(\Rightarrow(n+1) I_{n}=\frac{\pi}{4}-\int_{0}^{1} \frac{x^{n+1}}{1+x^{2}} d x\)----------(1)Put n → n + 2 in equation (1)
\(\Rightarrow(n+3) I_{n+2}=\frac{\pi}{4}-\int_{0}^{1} \frac{x^{n+3}}{1+x^{2}} d x\)--------(2)From equation(1) + (2)
⇒ (n +1)In + (n + 3) In+2 =
\(\frac{\pi }{2}-\frac{1}{n+2}\)⇒ on comparing with given equation
⇒an = n + 1, bn = n+ 3,
\(c_{n}=\frac{\pi }{2}-\frac{1}{n+2}\)
Question 69: Two particles A and B move from rest along a straight line with constant accelerations f and h respectively. If A takes m seconds more than B and describes n units more than that of B acquiring the same speed, then
- a. (f + h)m2 = fhn
- b. (f – fh)m2 = fhn
- c. (h – f)n = fhm2
- d. (1/2)(f + h)n =fhm2
Solution:
Answer: c
S + n = (1/2) f(t + m2)and S =(1/2)ht2 , V = ht
(1/2)ht2 +n= (1/2) f(t + m2) ……….(1)
Also V = 0 + ht = 0 + f(t +m) ⇒ t + m =(ht/f) (put in equation (1))
from equation (1)
\(\Rightarrow \frac{1}{2} \mathrm{ht}^{2}+\mathrm{n}=\frac{1}{2} \mathrm{f}\left(\frac{\mathrm{ht}}{\mathrm{f}}\right)^{2}\)\(\Rightarrow \mathrm{t}^{2}=\frac{2 \mathrm{nf}}{\mathrm{h}(\mathrm{h}-\mathrm{f})}\)Also,
\(\mathrm{ht}=\mathrm{f}(\mathrm{t}+\mathrm{m}) \Rightarrow \mathrm{t}^{2}=\frac{\mathrm{m}^{2} \mathrm{f}^{2}}{(\mathrm{h}-\mathrm{f})^{2}}\)Therefore,
\(\frac{2 \mathrm{nf}}{\mathrm{h}(\mathrm{h}-\mathrm{f})}=\frac{\mathrm{m}^{2} \mathrm{f}^{2}}{(\mathrm{h}-\mathrm{f})^{2}}\)\(\Rightarrow \mathrm{n}(\mathrm{h}-\mathrm{f})=\mid \frac{1}{2} \mathrm{f} \mathrm{hm}^{2}\)
Question 70: The area bounded by y = x + 1 and y = cos x and the x-axis, is
- a. 1 sq. unit
- b. (3/2) sq. unit
- c. (1/4) sq. unit
- d. (1/8) sq. unit
Solution:
Answer: b
\(Area=\frac{1}{2}\times 1\times 1+\int_{0}^{\pi /2}cosxdx\)\(=\frac{1}{2}+\left.\sin x\right|_{0} ^{\pi / 2}\)= (1/2)+1= 3/2
Question 71: Let x1, x2 be the roots of x2 – 3x + a = 0 and x3, x4 be the roots of x2 – 12x + b = 0. If x1< x2< x3< x4 and x1, x2, x3, x4 are in G.P. then ab equals
- a. 24/5
- b. 64
- c. 16
- d. 8
Solution:
Answer: b
Since x1 , x2, x3, x4 are in G.P, then
\(\frac{x_{3}+x_{4}}{x_{1}+x_{2}}=\frac{12}{3}\)\(\frac{Ar^{2}+Ar^{3}}{A+Ar}=4\)⇒r2 = 4 ⇒ x = 2 (since G.P is increasing)
⇒x1 + x2 = 3
⇒A + Ar = 3
⇒ A(3) = 3
⇒A = 1
Therefore, ab = x1 • x2 • x3 • x4 = 1.2. 4.8 = 64
Question 72: If q
- a. (2n + 1) (π/2) , n \(\epsilon I\)
- b.3nπ/2 , n \(\epsilon\)I
- c. nπ, n \(\epsilon\)I
- d. 2nπ, n \(\epsilon\)I
Solution:
Answer: a
Using rationalization
Question 73: Let A =
- a. 3, 0, 3
- b. 0, 3, 6
- c. 1, 0, –6
- d. 3, 3, 6
Solution:
Answer: b
(A – λI3) = 0
\(\begin{pmatrix} 3-\lambda & 0 &3 \\ 0& 3-\lambda &0 \\ 3& 0 & 3-\lambda \end{pmatrix}\)⇒ (3 – λ)3 – 9 (3 – λ) = 0
⇒ (3 – λ) [(3 – λ)2 – 32 = 0
⇒3 – λ = 0 or 3 – λ – 3 = 0 or 3 – λ +3 =0
⇒λ = 0, 3, 6
Question 74: Straight lines x – y = 7 and x + 4y = 2 intersect at B. Points A and C are so chosen on these two lines such that AB = AC. The equation of line AC passing through (2, –7) is:
- a. x – y – 9 = 0
- b. 23x + 7y + 3 = 0
- c. 2x – y – 11 = 0
- d. 7x – 6y – 56 = 0
Solution:
Answer: b
AB = AC (given)
∠ABC = ∠BCA
Let slope of AC is m.
m = -23/7, 1 (rejected)
Therefore, equation of line is 23x + 7y + 3 = 0
Question 75: Equation of a tangent to the hyperbola 5x2 – y2 = 5 and which passes through an external point (2, 8) is
- a. 3x – y + 2 = 0
- b. 3x + y – 14 = 0
- c. 23x – 3y – 22 = 0
- d. 3x – 23y + 178 = 0
Solution:
Answer: (a,c)
\(\frac{x^{2}}{1}-\frac{y^{2}}{5}=1\)Let the tangent of hyperbola
\(y=mx\pm \sqrt{a^{2}m^{2}-b^{2}}\)\(y=mx\pm \sqrt{m^{2}-5}\)……(1)Passes through (2, 8)
⇒ (8 – 2m) =
\(\pm \sqrt{m^{2}-5}\)On squaring both sides
⇒ (8 – 2m)2 = (m2 – 5)
⇒m = 3 or (23/3) (put in equation (1))
So equation of tangent is ⇒
\(y=3x\pm 2\)or\(y=\frac{23x}{3}\pm \frac{22}{3}\)⇒ y = 3x + 2, y = 3x – 2, 3y = 23x + 22, 3 y = 23x – 22
Video Lessons - WBJEE 2019 - Maths
WBJEE 2019 Maths Question Paper with Solutions
